02/22/2009, 10:53 PM
Gottfried Wrote:I see the noninvertibility of A for one single reason. First: A is decomposbale into two well known triangular matrices Binomial P and Stirling-kind2 S2 (factorially scaled).
Both factors P and S2 are invertible, so
\( A = S2 * P\sim \)
and formally hte inverse is possible
\( A^{-1} = P^{-1}\sim * S2^{-1} = P^{-1}\sim * S1 \)
where S1 is the matrix of Stirlingnumbers 1st kind, also factorially scaled.
The reason why A is not invertible, is that in \( P^{-1}\sim * S1 \) the dotproduct first row by second column is infinite and exactly gives zeta(1).
I think I somewhere read that for infinite matrices not even \( A(BC)=(AB)C \) is valid, perhaps exactly because one limit does not exist.
